Q124 / 150 · MHT CET 2026 (11 April Shift 1) · Physics
A particle of mass $m$ is moving in a circular path of constant radius $r$ such that its centripetal acceleration $a_c$ is varying with time $t$ as, $a_c = k^2 r t^2$. The power delivered to the particle by the forces acting on it is ($K$ = constant)
- A$m^2 k^2 r^2 t^2$
- Correct answerB$m k^2 r^2 t$
- C$2\pi m k^2 r^2 t$
- D$\dfrac{m k^4 r^2 t^5}{3}$
Solution
Given centripetal acceleration $a_c = k^2 r t^2$
Using the formula $a_c = \dfrac{v^2}{r}$:
$\dfrac{v^2}{r} = k^2 r t^2$
$v^2 = k^2 r^2 t^2$
$v = k r t$
The kinetic energy of the particle is given by:
$E = \dfrac{1}{2} m v^2$
$E = \dfrac{1}{2} m (k r t)^2 = \dfrac{1}{2} m k^2 r^2 t^2$
The total power delivered by all forces acting on the particle is equal to the rate of change of its kinetic energy:
$P = \dfrac{dE}{dt}$
$P = \dfrac{d}{dt} \left( \dfrac{1}{2} m k^2 r^2 t^2 \right)$
$P = \dfrac{1}{2} m k^2 r^2 (2t)$
$P = m k^2 r^2 t$
Using the formula $a_c = \dfrac{v^2}{r}$:
$\dfrac{v^2}{r} = k^2 r t^2$
$v^2 = k^2 r^2 t^2$
$v = k r t$
The kinetic energy of the particle is given by:
$E = \dfrac{1}{2} m v^2$
$E = \dfrac{1}{2} m (k r t)^2 = \dfrac{1}{2} m k^2 r^2 t^2$
The total power delivered by all forces acting on the particle is equal to the rate of change of its kinetic energy:
$P = \dfrac{dE}{dt}$
$P = \dfrac{d}{dt} \left( \dfrac{1}{2} m k^2 r^2 t^2 \right)$
$P = \dfrac{1}{2} m k^2 r^2 (2t)$
$P = m k^2 r^2 t$
Hence, the correct answer is option: 'B'.